For 0- which of the following is true

Let y=√(x) and two points A≡(α,√(α)), B≡(β,√(β)) ⇒d^2ydx^2= x^ 324 lt;0 so, downward concaving. From graph of y=√(x) : N is the mid point of AB; nbsp;⇒ N≡ ...

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Byju's Answer

Standard XII

Mathematics

Angle between Two Lines

For 0<α<β, wh...

Question

For $0 < α < β,$ which of the following is true

\sqrt{\frac{α + β}{2}} \leq \frac{\sqrt{α} + \sqrt{β}}{2}

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\sqrt{\frac{α + β}{2}} < \frac{\sqrt{α} + \sqrt{β}}{2}

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\sqrt{\frac{α + β}{2}} \geq \frac{\sqrt{α} + \sqrt{β}}{2}

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\sqrt{\frac{α + β}{2}} > \frac{\sqrt{α} + \sqrt{β}}{2}

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Solution

The correct option is D $\sqrt{\frac{α + β}{2}} > \frac{\sqrt{α} + \sqrt{β}}{2}$
Let $y = \sqrt{x}$
and two points $A \equiv (α, \sqrt{α}), B \equiv (β, \sqrt{β})$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = - \frac{x^{- \frac{3}{2}}}{4} < 0$
so, downward concaving.
From graph of $y = \sqrt{x} :$

$N$ is the mid point of $A B;$
$\Rightarrow N \equiv (\frac{α + β}{2}, \frac{\sqrt{α} + \sqrt{β}}{2})$
and $M \equiv (\frac{α + β}{2}, \sqrt{\frac{α + β}{2}})$
Clearly, Point $M$ lies above $N$
$∴ \sqrt{\frac{α + β}{2}} > \frac{\sqrt{α} + \sqrt{β}}{2}$

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For 0<α<β, which of the following is true

For $0 < α < β,$ which of the following is true